Answer:
1211 N.
Explanation:
Okay, we are given the following data or parameters or information in the question above;
=> "aluminum (E = 70 GPa) tube of length 8-m.
=> "The aluminum is used as a simply supported column carrying a 1.2 kN axial load"
The axial load to provide a factor of safety of 2 against buckling =[ (22/7)^2 × 224348 × 1000 × 70] ÷ (8 × 1000)^2 × 2.
The axial load to provide a factor of safety of 2 against buckling = 1211 N.
A steam turbine receives 8 kg/s of steam at 9 MPa, 650 C and 60 m/s (pressure, temperature and velocity). It discharges liquid-vapor mixture with a quality of 0.94 at a pressure of 325 kPa and a velocity of 15 m/s. In addition, there is heat transfer from the turbine to the surroundings for 560 kW. Find the power produced by the turbine and express it in kW?
Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
[tex]Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ][/tex]
Which gives;
[tex]560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ][/tex]
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
[tex]- \dot{W }[/tex] = -22749.1856 - 560 = -23309.1856 kJ
[tex]\dot{W }[/tex] = 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Asphalt at 120 F, considered to be a Newtonian fluid with a viscosity 80000 times that of water and a specific gravity of 1.09, flow through a pipe of diameter 2.0 in. If the pressure gradient is 1.6 psi/ft determine the flowrate assuming the pipe is (a) horizontal; (b) vertical with flow up.
Answer:
a) Flow rate if the pipe is horizontal, Q = 4.69 * 10⁻³ ft³/s
b) Flow rate if the pipe is vertical, Q = 3.30 * 10⁻³ ft³/s
Explanation:
From the BI table, dynamic viscosity of water at 120°F is:
[tex]\mu_{H_2O} = 1.164 * 10^{-5} lb.s/ft^2[/tex]
Pressure gradient, [tex]\frac{\delta p}{\delta x} = 1.6 psi/ft[/tex]
Pipe Diameter, D = 2 in = 2/12 ft = 0.167 ft
Dynamic viscosity of asphalt at 120°F:
[tex]\mu = v \mu_{H_2O}\\\mu = 80000 * 1.164 * 10^{-5}\\\mu = 0.9312 lb-s/ft^2[/tex]
Specific weight of asphalt:
[tex]\gamma = SG \gamma_{H_2O}\\\gamma = (1.09)(62.4)\\\gamma = 68.016 lb/ft^3[/tex]
Flow rate, Q, of the asphalt when the pipe is in horizontal position assuming that the flow is laminar:
Note that if the pipe is horizontal, θ = 0°
[tex]Q = \frac{\pi D^4}{128 \mu} [(\frac{\delta p }{\delta x}) - \gamma sin \theta]\\\\Q = \frac{\pi 0.167^4}{128 * 0.9312} [(1.6*144) - 68.016 sin (0)]\\\\Q = 4.69 * 10^{-3} ft^3 / s[/tex]
b) Flow rate assuming the pipe is vertical:
At vertical pipe position, θ = 90°
[tex]Q = \frac{\pi D^4}{128 \mu} [(\frac{\delta p }{\delta x}) - \gamma sin \theta]\\\\Q = \frac{\pi 0.167^4}{128 * 0.9312} [(1.6*144) - 68.016 sin (90)]\\\\Q = 3.30 * 10^{-3} ft^3 / s[/tex]
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a stress state greater than 200 N/mm2. If the linkage is to be constructed from solid round stock, what is the minimum required diameter?
Answer:
minimum required diameter of the steel linkage is 3.57 mm
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = [tex]2*10^{8} N/m^{2}[/tex]
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/([tex]2*10^{8}[/tex]) = [tex]10^{-5}[/tex] m^2
recall that area = [tex]\pi d^{2} /4[/tex]
[tex]10^{-5}[/tex] = [tex]\frac{3.142*d^{2} }{4}[/tex] = [tex]0.7855d^{2}[/tex]
[tex]d^{2} = \frac{10^{-5} }{0.7855}[/tex] = [tex]1.273*10^{-5}[/tex]
[tex]d = \sqrt{1.273*10^{-5} }[/tex] = [tex]3.57*10^{-3}[/tex] m = 3.57 mm
maximum diameter of the steel linkage d = 3.57 mm
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. If the inside surface temperature is 1450 °C and outer surface temperature is 90°C, estimate the temperature between layers and calculate the heat loss in KJ/h-m2. Assume k for fire brick 4 KJ/m-h°C, k for common brick= 2.8 KJ/m-h°C, k for 85% magnesia = 0.25 KJ/m-h°C and k for steel 240 KJ/m-h°C, k
Answer:
fire brick / common brick : 1218 °Ccommon brick / magnesia : 1019 °Cmagnesia / steel : 90.06 °Cheat loss: 4644 kJ/m^2/hExplanation:
The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...
R = d/k
so the thermal resistances of the layers of furnace wall are ...
R₁ = 0.200/4 = 0.05 °C·m²·h/kJ
R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ
R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ
R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ
So, the total thermal resistance is ...
R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ
__
The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)
__
The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:
fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C
so, the fire brick interface temperature at the common brick is ...
1450 -232 = 1218 °C
For the next layers, the interface temperatures are ...
common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C
magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C
_____
Comment on temperatures
Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.
The equilibrium fraction of lattice sites that are vacant in electrum (a silver gold alloy) at 300K C is 9.93 x 10-8. There are 5.854 x 1015 vacancies/cm3.The atomic mass of electrum is 146.08 g/mol.Avagadro number 6.02210^23 atom/mol.Calculate the density of electrum in g/cm3.
Answer:
the density of the electrum is 14.30 g/cm³
Explanation:
Given that:
The equilibrium fraction of lattice sites that are vacant in electrum = [tex]9.93*10^{-8}[/tex]
Number of vacant atoms = [tex]5.854 * 10^{15} \ vacancies/cm^3[/tex]
the atomic mass of the electrum = 146.08 g/mol
Avogadro's number = [tex]6.022*10^{23[/tex]
The Number of vacant atoms = Fraction of lattice sites × Total number of sites(N)
[tex]5.854*10^{15}[/tex] = [tex]9.93*10^{-8}[/tex] × Total number of sites(N)
Total number of sites (N) = [tex]\dfrac{5.854*10^{15}}{9.93*10^{-8}}[/tex]
Total number of sites (N) = [tex]5.895*10^{22}[/tex]
From the expression of the total number of sites; we can determine the density of the electrum;
[tex]N = \dfrac{N_A \rho _{electrum}}{A_{electrum}}[/tex]
where ;
[tex]N_A[/tex] = Avogadro's Number
[tex]\rho_{electrum} =[/tex] density of the electrum
[tex]A_{electrum}[/tex] = Atomic mass
[tex]5.895*10^{22} = \dfrac{6.022*10^{23}* \rho _{electrum}}{146.08}[/tex]
[tex]5.895*10^{22} *146.08}= 6.022*10^{23}* \rho _{electrum}[/tex]
[tex]8.611416*10^{24}= 6.022*10^{23}* \rho _{electrum}[/tex]
[tex]\rho _{electrum}=\dfrac{8.611416*10^{24}}{6.022*10^{23}}}[/tex]
[tex]\mathbf{ \rho _{electrum}=14.30 \ g/cm^3}[/tex]
Thus; the density of the electrum is 14.30 g/cm³
Describe experimental factors that could be modified, and unalterable properties of materials used.
Answer:
a. mechanical properties
b. thermal properties
c. chemical properties
d. electical properties
e. magnetic properties
Explanation:
a. The mechanical properties of a material are those properties that involve a reaction to an applied load.The most common properties considered are strength, ductility, hardness, impact resistance, and fracture toughness, elasticity, malleability, youngs' modulus etc.
b. Thermal properties such as boiling point , coefficient of thermal expansion , critical temperature , flammability , heat of vaporization , melting point ,thermal conductivity , thermal expansion ,triple point , specific heat capacity
c. Chemical properties such as corrosion resistance , hygroscopy , pH , reactivity , specific internal surface area , surface energy , surface tension
d. electrical properties such as capacitance , dielectric constant , dielectric strength , electrical resistivity and conductivity , electric susceptibility , nernst coefficient (thermoelectric effect) , permittivity etc.
e. magnetic properties such as diamagnetism, hysteresis, magnetostriction , magnetocaloric coefficient , magnetoresistance , permeability , piezomagnetism , pyromagnetic coefficient
A long corridor has a single light bulb and two doors with light switch at each door.
Design logic circuit for the light; assume that the light is off when both switches are
in the same position.
Answer:
Light = A xor B
Explanation:
If switches A and B produce True or False, then Light will be True for ...
Light = A xor B
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the specimen fractured at a load of 4250 N when the distance between support points was 44 mm. Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm in length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 44 mm
Answer:
F = 8849 N
Explanation:
Given:
Load at a given point = F = 4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ[tex]_{fs} = F_{f} L / \pi R^{3}[/tex]
σ = FL / π R³
Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³
= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³
= 187 / 3.141593 (0.0056)³
= 338943767.745358
= 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
[tex]F_{f}[/tex] = 2 σ[tex]_{fs}[/tex] d³/3 L
F = 2σd³/3L
= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)
= 146016 / 125 / 3 ( 44 x 1 / 1000 )
= ( 146016 / 125 ) / (3 ( 11 / 250 ))
= 97344 / 11
F = 8849 N
Fluid flow in a pipe, the pressure decrease due to throttle valve from 10 bar to 100 kpa if the specific volume of fluid increase from 0.3 m3 /kg to 1.8 m3/kg. Find the change in internal energy during the process
Answer:
Fluid flow in a pipe, the pressure decrease due to throttle valve from 10 bar to 100 kpa if the specific volume of fluid increase from 0.3 m3 /kg to 1.8 m3/kg. Find the change in internal energy during the process
Explanation:
hope it will helps you
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I send
Answer:
32000 bits/seconds
Explanation:
Given that :
there are 16 signal combinations (states) = 2⁴
bits n = 4
and a baud rate (number of signals/second) = 8000/second
Therefore; the number of bits per seconds can be calculated as follows:
Number of bits per seconds = bits n × number of signal per seconds
Number of bits per seconds = 4 × 8000/second
Number of bits per seconds = 32000 bits/seconds
help mhee why are you u an enigner
Answer:
help me why are you an enginer
Explanation:
because lives
Select True/False for each of the following statements regarding aluminum / aluminum alloys: (a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor. (b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum. (c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic. (d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation. (e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
Explanation:
(a) Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.
False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their micro structures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.
True.
(c) Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.
False, aluminium is stable at high temperatures and does not oxidizes.
(d) Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.
False,pure aluminium is not resistant to the creep deformation.
(e) The relatively low melting point of aluminum is often considered a significant limitation for high-temperature structural applications.
False.
In this exercise, we have to analyze the statements that deal with aluminum and its properties, thus classifying it as true or false:
A) False
B) True
C) False
D) False
E) True
Analyzing the statements we can classify them as:
(a) For this statement we can say that it is False, aluminium is not susceptible to any corrosion by the presence of water vapor.
(b) For this statement we can say that it is True.
(c) For this statement we can say that it is False, aluminium is stable at high temperatures and does not oxidizes.
(d) For this statement we can say that it is False, pure aluminium is not resistant to the creep deformation.
(e) For this statement we can say that it is True.
See more about aluminum properties at brainly.com/question/12867973
An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression when the lintel carries a total uniformly distributed load of 10000 lb on a simple span of 6 ft. Also, calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange
Answer:
hello your question lacks some information attached is the complete question
A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in
(ii) maximum bending stress in compression = 0.7413*10^6 Ib-in
B) (i) The average shear stress at the neutral axis = 0.7904 *10 ^5 psi
(ii) Average shear stress at the web = 18.289 * 10^5 psi
(iii) Average shear stress at the Flange = 1.143 *10^5 psi
Explanation:
First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)
A) Calculate the maximum bending stress in tension and compression
lintel load = 10000 Ib
simple span = 6 ft
( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS
I = 53.54
i) The maximum bending stress (fb) in tension=
= [tex]\frac{M_{mm}Y }{I}[/tex] = [tex]\frac{6.48 * 10^6 * 2.375}{53.54}[/tex] = 0.287 * 10^6 Ib-in
ii) The maximum bending stress (fb) in compression
= [tex]\frac{M_{mm}Y }{I}[/tex] = [tex]\frac{6.48 *10^6*(8.5-2.375)}{53.54}[/tex] = 0.7413*10^6 Ib-in
B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange
i) The average shear stress at the neutral axis
V = [tex]\frac{wL}{2}[/tex] = [tex]\frac{1000*6*12}{2}[/tex] = 3.6*10^5 Ib
Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - [tex]\frac{0.5}{2}[/tex] ) * [tex]\frac{(2.375 - (\frac{0.5}{2} ))}{2}[/tex]
= 5.878 in^3
t = VQ / Ib = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi
ii) Average shear stress at the web ( value gotten from the shear stress at the flange )
t = 1.143 * 10^5 * (8 / 0.5 ) psi
= 18.289 * 10^5 psi
iii) Average shear stress at the Flange
t = VQ / Ib = [tex]\frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}[/tex]
= 1.143 *10^5
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released.
Required:
a. If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
b. If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
c. Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk?
Answer:
a) The mass moves a distance of 0.625 m up the slope before coming to rest
b) The distance moved by the mass when it is connected to the spring is 0.6 m
c) [tex]\mu = 0.206[/tex]
Explanation:
Spring constant, k = 70 N/m
Compression, x = 0.50 m
Mass placed at the free end, m = 2.2 kg
angle, θ = 41°
Potential Energy stored in the spring, [tex]PE= 0.5 kx^2[/tex]
[tex]PE = 0.5 * 70 * 0.5^2\\PE = 8.75 J[/tex]
According to the principle of energy conservation
PE = mgh
8.75 = 2.2 * 9.8 * h
h = 0.41
If the mass moves a distance d from the spring
sin 41 = h/d
sin 41 = 0.41/d
d = 0.41/(sin 41)
d = 0.625 m
The mass moves a distance of 0.625 m up the slope before coming to rest
b) If the mass is attached to the spring
According to energy conservation principle:
Initial PE of spring = Final PE of spring + PE of block
[tex]0.5kx_1^2 = 0.5kx_2^2 + mgh\\x_2 = d - x_1 = d - 0.5\\h = d sin 41\\0.5*70*0.5^2 = 0.5*70*(d-0.5)^2 + 2.2*9.8*d*sin41\\8.75 = 35(d^2 - d + 0.25) + 14.15d\\8.75 = 35d^2 - 35d + 8.75 + 14.15d\\35d^2 = 20.85d\\d = 0.6 m[/tex]
The distance moved by the mass when it is connected to the spring is 0.6 m
3) The spring potential is converted to increased PE and work within the system.
mgh = Fd + 0.5kx²...........(1)
d = x , h = dsinθ
kinetic friction force , F = μmgcosθ
mgdsinθ + μmg(cosθ)d = 0.5kd²
mgsinθ + μmgcosθ = 0.5kd
sinθ + μcosθ = kd/(2mg)
[tex]\mu = \frac{\frac{kd}{2mg} - sin\theta}{cos\theta} \\\\\mu = \frac{\frac{70*0.5}{2*2.2*9.8} - sin41}{cos41} \\\\\mu = 0.206[/tex]
Answer:
A) l = 0.619m
B) l = 0.596m
C) μ = 0.314
Explanation:
The data given is:
k = 70 N/m
x = 0.5 m
m = 2.2 kg
θ = 41°
(FIGURES FOR EACH PART ARE ATTACHED AT THE BOTTOM. CONSULT THEM FOR BETTER UNDERSTANDING)
Part A
Gain in Gravitational Potential Energy = Loss in Elastic Potential Energy
mgh = (1/2)kx²
(2.2)(9.8)h = (1/2)(70)(0.5)²
h = 0.406 m
sinθ = h/l
l = h / sinθ
l = 0.406/sin41
l = 0.619m
Part B
Loss in Elastic Potential Energy in compressed spring = Gain in Gravitational Potential Energy + Gain in Elastic Potential Energy in stretched spring
(1/2)kx² = mgh + (1/2)k(l - 0.5)²
(1/2)(70)(0.5)² = (2.2)(9.8)(l·sin41)) + (1/2)(70)(l² - l + 1/4)
8.75 = 14.15(l) + 35(l²) - 35(l) + 8.75
35(l²) -20.85(l) = 0
l = 0.596m
Part C
Loss in Elastic Potential Energy = Gain in Gravitational Potential Energy + Work done against friction
(1/2)kx² = mgh + Fd
(1/2)kx² = mg(dsinθ) + μRd
(1/2)kx² = mg(dsinθ) + μ(mg · cosθ)d
(1/2)kx² = mgd (sinθ + μ(cosθ))
(1/2)(70)(0.5)² = (2)(9.8)(0.5) (sin41 + μcos41)
8.75 = 6.43 + 7.4μ
μ = 0.314
Which of the following are the main psychological domains?
Answer:
Domain 1: Biological (includes neuroscience, consciousness, and sensation) Domain 2: Cognitive (includes the study of perception, cognition, memory, and intelligence) Domain 3: Development (includes learning and conditioning, lifespan development, and language) i hope this helps you.
Consider a potato being baked in an oven that is maintained at a constant temperature. the temperature of the potato is observed to rise by 5c during the first minute. Will the temperature rise during the second minute be less than, equal to, or more than 5∘C? Why?
Answer:
The rise in temperature will be less than 5 °C in the second minute.
Explanation:
According to heat conduction law, the rate of heating is proportional to the temperature difference or temperature gradient. The temperature gradient is what drives heat to move from a hotter body at a higher temperature gradient to a colder body at a lower temperature gradient. For the potato, the initial first minute raises the temperature to 5 °C, consequently reducing the temperature gradient between the potato and the heating element in the oven. This reduced temperature gradient means that the rate at which it will conduct heat in the second minute will be lesser than that at the first minute. This will continue till the potato and the heating element are at the same temperature, at which no temperature gradient will exist between them; stopping heat transfer between them.
A nail gun operates using pressurized air, which is supplied through a 1/4-in diameter hose. The gun requires 75 psi to operate with a 2.6 ft3 /min airflow. If the air compressor develops 90 psi, determine the maximum allowable length of hose that can be used for its operation. Assume incompressible flow and a smooth pipe. Take =0.000238 slug/ft3 and = 1.6×10-4 ft2 /s.
Answer:
If we assumed flow was laminar, L = 1840 m
But in reality, this flow is in the turbulent region and L = 0.00000304 m = (1.197 × 10⁻⁴) inch
Explanation:
We first check the region of flow of the fluid by computing the Reynolds number
Re = (ρvD/μ)
Listing all the parameters and converting to SI units
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
v = velocity of flow = (Q/A)
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
A = Cross sectional Area of the pipe = (πD²/4)
D = Pipe diameter = (1/4) inch = 0.00635 m
A = (π×0.00635²/4) = 0.00003167 m²
v = (0.0012271/0.00003167) = 38.746 m/s
μ = dynamic viscosity of the fluid = (Kinematic viscosity) × (density)
Kinematic viscosity = (1.6 × 10⁻⁴) ft²/s = (1.486 × 10⁻⁵) m²/s
μ = (1.486 × 10⁻⁵) × (0.123) = 0.0000018278 = (1.83 × 10⁻⁶) Pa.s
Re = (0.123×38.746×0.00635)/(1.83 × 10⁻⁶) = 16,556.824214903
This Reynolds number is in the turbulent flow region.
From Hagen-Poiseulle equation, the volumetric flowrate for laminar and turbulent flow is given as
Q = (πD⁴ΔP)/(128μL) for laminar flow
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵ for turbulent flow
Since our flow is turbulent
ΔP = (0.241 × ρ⁰•⁷⁵ × μ⁰•²⁵ × L × Q¹•⁷⁵)/D⁴•⁷⁵
L = (ΔP.D⁴•⁷⁵)/(0.241 ρ⁰•⁷⁵ μ⁰•²⁵ Q¹•⁷⁵)
Listing all the parameters and converting to SI units
L = Length of the pipe = ?
ΔP = Pressure drop across the flow channel = 90 - 75 = 15 psi = 103421.4 Pa
D = Pipe diameter = (1/4) inch = 0.00635 m
ρ = density of the fluid = 0.000238 slug/ft³ = 0.123 kg/m³
μ = dynamic viscosity of the fluid = (1.83 × 10⁻⁶) Pa.s
Q = volumetric flowrate of the air = 2.6 ft³/min = 0.0012271 m³/s
L = (0.123 × 0.00635⁴•⁷⁵) ÷ (0.241 × 0.123⁰•⁷⁵ × 0.0000018278⁰•²⁵ × 0.0012271¹•⁷⁵)
L = (4.4986 × 10⁻¹²) ÷ (1.481 × 10⁻⁶)
L = 0.0000030378 m = 0.00000304 m = (1.197 × 10⁻⁴) inch
If we assumed that flow was laminar
Q = (πD⁴ΔP)/(128μL)
L = (πD⁴ΔP)/(128μQ)
L = (π × 0.00635⁴ × 103421.4) ÷ (128 × 0.0000018278 × 0.0012271)
L = (0.0005282691) ÷ (0.0000002871)
L = 1840 m
Hope this Helps!!!
Hot coffee is contained in a cylindrical thermos bottle that is of length L-0.3m and is lying on its side (horizontally). The coffee container consists of a glass flask of diameter DI-7 cm sperated from an aluminum housing of diameter D2-0.08 m by air at atmospheric pressure. If the inner and outer temperatures are 75oC and 35 oC respectively, what is the heat loss from the coffee due to convection?
Note: The diagram attached below is the completion of the given question.
Also calculate the heat loss from the coffee due to radiation and the total heat loss
Answer:
[tex]Q_{conv} = 16.04 W\\Q_{rad} = 3.20 W\\H_{loss} = 19.24 W[/tex]
Explanation:
Glass diameter of the coffee container, [tex]D_1 = 7 cm = 0.07 m[/tex]
Glass radius of the coffee container, r₁ = 0.07/2 = 0.035 m
Diameter of the aluminium housing, [tex]D_2 = 0.08 m[/tex]
Radius of the aluminium housing, r₂ = 0.08/2 = 0.04 m
Inner temperature, T₁ = 75°C = 348 K
Outer temperature, T₂ = 35°C = 308 K
[tex]\epsilon_1 = 0.25\\\epsilon_2 = 0.25[/tex]
[tex]T_f = \frac{T_1 + T_2}{2} \\T_f = \frac{348+308}{2} \\T_f = 328 K[/tex]
At [tex]T_f = 328 K[/tex], P = 1 atm
k = 0.0284 w/m-k, v = 23.74 * 10⁻⁶, [tex]\alpha = 26.6 * 10^{-6}[/tex], pr = 0.703, [tex]\beta = 3.05 * 10^{-3} K^{-1}[/tex]
[tex]H_{loss} = Q_{rad} + Q_{conv}[/tex]
[tex]Q_{conv} = \frac{2 \pi k L (T_1 - T_2)}{ln(\frac{r_2}{r_1} )} \\\\Q_{conv} = \frac{2 \pi* 0.0284 * 0.3 (75 - 35)}{ln(\frac{0.04}{0.035} )} \\Q_{conv} = 16.04 W[/tex]
[tex]Q_{rad} = \frac{\sigma (\pi D_1 L) (T_1^4 - T_2^4)}{\frac{1}{\epsilon_1 } + \frac{1 - \epsilon_2}{\epsilon_2} (\frac{\gamma_1}{\gamma_2}) }[/tex]
[tex]Q_{rad} = \frac{5.67*10^{-8}(\pi * 0.07*0.3) (348^4 - 308^4)}{\frac{1}{0.25 } + \frac{1 - 0.25}{0.25} (\frac{0.035}{0.04}) } \\\\Q_{rad} = 3.20 W[/tex]
[tex]H_{loss} = 3.20 + 16.04\\H_{loss} = 19.24 W[/tex]
Calculate the resistance using Voltage and current, again using voltage and power, again using current and power, and again using R1 and R2 recording the calculations for Run 3 rows 41-56
Answer:
R = V / I , R = V² / P, R = P / I²
Explanation:
For this exercise let's use ohm's law
V = I R
R = V / I
Electric power is defined by
P = V I
ohm's law
I = V / R
we substitute
P = V (V / R)
P = V² / R
R = V² / P
the third way of calculation
P = (i R) I
P = R I²
R = P / I²
A vehicle experiences hard shifting. Technician A says that the bell housing may be misaligned. Technician B says that incorrect oil may have been put in the transmission. Who is correct? Group of answer choices
Answer:
Technician B
Explanation:
Vehicle hard shifting is a situation whereby the vehicle faces difficulty or shakes when changing gears/speed.
Actually technician B is correct, the primary reason for hard shifting is low level of transmission fluid, hard shifts can also be caused by excessive line pressure due to a clog or malfunctioning shift solenoid.
Consider two double-pipe counter-flow heat exchangers that are identical except that one is twice as long as the other. Is the longer heat exchanger more likely to have a higher effectiveness
Answer:
Both pipes will have the same effectiveness
Explanation:
Heat exchanger effectiveness is defined as the ratio of the actual amount of heat transferred to the maximum possible amount of heat that could be transferred with an infinite area.
The actual amount of heat transferred for counter-flow heat exchangers is given as;
[tex]q = C_h(T_h,_i -T_h,_o)[/tex]
where;
Ch is the specific heat capacity
Th,i is the inlet temperature of the hot fluid
Th,o is the outlet temperature of the hot fluid
The maximum possible amount of heat that could be transferred with an infinite area, is given as;
[tex]q_{max} = C_{min}(T_h,_i-T_c,_i)[/tex]
where;
Cmin is the minimum heat transfer coefficient for hot and cold fluid
Heat exchanger effectiveness is calculated as;
[tex]\epsilon = \frac{q}{q_{max}}[/tex]
From the formula above, Heat exchanger effectiveness is independent of the pipe length.
Therefore, both pipes will have the same effectiveness
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 10 MPa, 450°C, and 80 m/s, and the exit conditions are 10 kPa, 92% quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s.
Determine:
(a) the change in kinetic energy
(b) the power output
(c) the turbine inlet area
Answer:
a) The change in Kinetic energy, KE = -1.95 kJ
b) Power output, W = 10221.72 kW
c) Turbine inlet area, [tex]A_1 = 0.0044 m^2[/tex]
Explanation:
a) Change in Kinetic Energy
For an adiabatic steady state flow of steam:
[tex]KE = \frac{V_2^2 - V_1^2}{2} \\[/tex].........(1)
Where Inlet velocity, V₁ = 80 m/s
Outlet velocity, V₂ = 50 m/s
Substitute these values into equation (1)
[tex]KE = \frac{50^2 - 80^2}{2} \\[/tex]
KE = -1950 m²/s²
To convert this to kJ/kg, divide by 1000
KE = -1950/1000
KE = -1.95 kJ/kg
b) The power output, w
The equation below is used to represent a steady state flow.
[tex]q - w = h_2 - h_1 + KE + g(z_2 - z_1)[/tex]
For an adiabatic process, the rate of heat transfer, q = 0
z₂ = z₁
The equation thus reduces to :
w = h₁ - h₂ - KE...........(2)
Where Power output, [tex]W = \dot{m}w[/tex]..........(3)
Mass flow rate, [tex]\dot{m} = 12 kg/s[/tex]
To get the specific enthalpy at the inlet, h₁
At P₁ = 10 MPa, T₁ = 450°C,
h₁ = 3242.4 kJ/kg,
Specific volume, v₁ = 0.029782 m³/kg
At P₂ = 10 kPa, [tex]h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg[/tex], x₂ = 0.92
specific enthalpy at the outlet, h₂ = [tex]h_1 + x_2 h_{fg}[/tex]
h₂ = 3242.4 + 0.92(2392.1)
h₂ = 2392.54 kJ/kg
Substitute these values into equation (2)
w = 3242.4 - 2392.54 - (-1.95)
w = 851.81 kJ/kg
To get the power output, put the value of w into equation (3)
W = 12 * 851.81
W = 10221.72 kW
c) The turbine inlet area
[tex]A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2[/tex]
A particular brand of paint covers 340 square feet per gallon . Write a program in C++ to determine and report approximately how many gallons of paint will be needed to paint two coats on a wooden fence that is 6 feet high and 100 feet long .
Answer:
#include <iostream>
using namespace std;
// named constant to give mnemonic name to "magic number"
const float SqFtPerGal = 350.0f;
// the function main() is always the entry point of the application
int main()
{
float length, width, area, paint; // to hold user values and results
// prompt user for length and width of wall
cout << "Enter length of wall : " << flush;
cin >> length;
cout << "Enter width of wall : " << flush;
cin >> width;
// calculate area and amount of paint needed
area = length * width;
paint = area / SqFtPerGal;
// output results with reasonable text
cout << "You need " << paint << " gallons of paint to cover "
<< area << " square feet of wall." << endl;
// program stops executing when it returns from main(), 0 means O.K.
return 0;
}
Explanation:
Identify the correct statements in the context of friction factors of laminar and turbulent flows
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces
b) In turbulent flow, the tubes with rough surfaces have much lower friction factors than the tubes with smooth surfaces.
c) In laminar flow, the friction factor is dependent on the surface roughness
d) In laminar flow, the friction factor is independent of the surface roughness.
Answer:
a) In turbulent flow, the tubes with rough surfaces have much higher friction factors than the tubes with smooth surfaces.
Explanation:
Turbulent flow is a type of fluid flow in which fluid will undergo irregular fluctuations. The tubes with rough surfaces have higher friction factors than the tubes with smooth surfaces. In laminar flow the effect of effect of surface roughness is negligible on friction factors.
Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 10, and the inlet temperature for each turbine stage is 1400 K. The pressure ratios across each turbine stage are equal. The turbine stages and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k= 1.4.
Calculate:
a. the thermal efficiency of the cycle.
b. the back work ratio.
c. the net power developed, in kW.
Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
[tex]T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}[/tex]
T₁ = 300 K
[tex]\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }[/tex]
[tex]T_{2s}[/tex] = 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
[tex]\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{5s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₅ = T₄ + ([tex]T_{5s}[/tex] - T₄)/[tex]\eta _t[/tex] = 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + [tex]\epsilon _{regen}[/tex](T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
[tex]\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }[/tex]
[tex]T_{7s}[/tex] = 1400×[tex]\left( 1/\sqrt{10} \right)^{\dfrac{0.4}{1.4} }[/tex] = 1007.6 K
T₇ = T₆ + ([tex]T_{7s}[/tex] - T₆)/[tex]\eta _t[/tex] = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. [tex]W_{net \ out}[/tex] = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. [tex]bwr = \dfrac{W_{c,in}}{W_{t,out}}[/tex]
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters at a temperature and exits at 284 C. The pipe is smooth and its length is 10 m. temperature is 25 ° C. Since the smear coefficient of the pipe surface is given as 0.8; a-) Indoor and outdoor convection coefficients (W / m2K), b-) Heat loss from the pipe to the environment (W), c-) The surface temperature of the pipe (C), d-) Calculate the required fan control (W) and interpret the results.
Answer:
a) [tex]h_c = 0.1599 W/m^2-K[/tex]
b) [tex]H_{loss} = 5.02 W[/tex]
c) [tex]T_s = 302 K[/tex]
d) [tex]\dot{Q} = 25.125 W[/tex]
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature, [tex]T_a = 25^0 C = 298 K[/tex]
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
[tex]\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec[/tex]
Rate of heat transfer,
[tex]\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W[/tex]
a) To calculate the convection coefficient relationship for heat transfer by convection:
[tex]\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K[/tex]
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe, [tex]k_c = 0.8[/tex]
[tex]\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K[/tex]
b) Heat loss from the pipe to the environment:
[tex]H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W[/tex]
d) The required fan control power is 25.125 W as calculated earlier above
A closed, 5-m-tall tank is filled with water to a depth of 4 m. The top portion of the tank is filled with air which, as indicated by a pressure gage at the top of the tank, is at a pressure of 20 kPa. Determine the pressure that the water exerts on the bottom of the tank.
Answer:
The pressure that the water exerts on the bottom of the tank is 59.2 kPa
Explanation:
Given;
height of tank, h = 5m
height of water in the tank, [tex]h_w[/tex] = 4m
pressure at the top of the tank, [tex]P_{top}[/tex] = 20 kPa
The pressure exerted by water at the bottom of the tank is the sum of pressure on water surface and pressure due to water column.
[tex]P_{bottom} = \gamma h + P_{top}\\\\P_{bottom} = (9.8*10^3*4 \ \ + \ 20*10^3)Pa\\\\P_{bottom} = 59200 \ Pa\\\\P_{bottom} = 59.2 \ kPa[/tex]
Therefore, the pressure that the water exerts on the bottom of the tank is 59.2 kPa
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz. Determine the tube thickness that should be used if the shearing stress is not to exceed 60 MPa.
Answer:
25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
Explanation:
Given data
steel tube : outer diameter = 50-mm
power transmitted = 100 KW
frequency(f) = 34 Hz
shearing stress ≤ 60 MPa
Determine tube thickness
firstly we calculate the ; power, angular velocity and torque of the tube
power = T(torque) * w (angular velocity)
angular velocity ( w ) = 2[tex]\pi[/tex]f = 2 * [tex]\pi[/tex] * 34 = 213.71
Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s
next we calculate the inner diameter using the relation
[tex]\frac{J}{c_{2} } = \frac{T}{t_{max} }[/tex] = 467.92 / (60 * 10^6) = 7.8 * 10^-6 m^3
also
c2 = (50/2) = 25 mm
[tex]\frac{J}{c_{2} }[/tex] = [tex]\frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} )[/tex] = [tex]\frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1} ) ][/tex]
therefore; 0.025^4 - [tex]c^{4} _{1}[/tex] = 0.050 / [tex]\pi[/tex] (7.8 *10^-6)
[tex]c^{4} _{1}[/tex] = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)
39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8
[tex]c_{1} = \sqrt[4]{26.66 * 10^{-8} }[/tex] =
THE TUBE THICKNESS
[tex]c_{2} - c_{1}[/tex] = 25 - [tex]\sqrt[4]{26.66*10^{-8} }[/tex] mm
What information is needed to set up sales tax in QuickBooks Online for a client who only does business in their home state?
Answer:
Their company address
When their last tax period started
How often they have to file a tax return
When they started collecting sales tax for the agency
Explanation:
For setup the sales tax information in Quickbooks online for a client who only does business in their home state, we need these information which are given below:
1. Their company addresses
2. Last Tax Time period started
3. How frequently they filed the tax return
3. when they begin to received sales tax
Therefore all the other options are not valid. Hence, ignored it
Question 1 The first choice for how to reduce or eliminate a hazard is: a) Engineering controls b) Workplace controls c) Personal protective equipment d) Administrative controls
Answer:
a) Engineering controls.
Explanation:
Hazard can be defined as any agent or source that has the potential to constitute danger and cause harm, damage or adverse bodily injuries, health effects on a vulnerable individual, property or group of people.
Generally, can be classified into various categories such as mechanical, biological, chemical, physical, psychosocial and ergonomic hazard. These hazards are either human induced or natural.
Some examples of hazard are radiation, fire, flood, chemicals, drought, vapor or steam, exposed live wire, dust particles, electrical circuits and equipments etc.
The first choice for how to reduce or eliminate a hazard is engineering controls. Engineering controls of hazards involves the process of protecting, shielding or guarding individuals by eliminating the agent of hazards or through the use of barriers between the hazard and the vulnerable individual or group of people.
Basically, engineering controls when properly designed, maintained and used effectively would help to mitigate hazards and keep the work environment relatively safe for workers.
Examples of engineering controls are;
1. Ventilation systems.
2. Machine or equipment guards.
3. Radiation shields.
4. Safety interlocks.
5. Sound dampening equipments.